I'm asked to represent the function $\displaystyle \frac{2 x}{10 + x}$ as a power series $f(x) = \displaystyle \sum_{n=0}^\infty c_n x^n$
I found this to be $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{5 \cdot 10^n} x^{n+1}$
The question is asking for $\ c_0, c_1, c_2, c_3,$ and $\ c_4$
I entered $\displaystyle c_0 = \frac{1}{5}, c_1 = \frac{-1}{50}, c_2 = \frac{1}{500}, c_3 = \frac{-1}{5000},$ and $\displaystyle c_4 = \frac{1}{50000}$ but I am told this is incorrect.
Can someone please explain what I'm doing wrong? Thanks!
UPDATE: using another formula I was able to find $\displaystyle c_n = \frac{f^{n}(a)}{n!}$ so $\displaystyle c_0 = 0, c_1 = \frac{1}{5}, c_2 = \frac{-1}{50},$ etc.. which are the correct answers. So I'm still confused.. Assuming I calculated the power series representation correctly to get $\ c_0$ don't you just let n=0 and take the coefficient of the resulting term?
UPDATE: $\ c_n$ is actually the coefficient for the $\ x^n$th term. So $\ c_0$ is the constant term. This helped me for another question I had where the terms were $\ c_0 + c_2x^2 + c_4x^4 \cdots $ and I didn't understand why the coefficient $\ c_1$ was 0
Best Answer
You were asked to find a series in the form $$\displaystyle \sum_{n=0}^\infty c_n x^n$$ but your answer is $$\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{5 \cdot 10^n} x^{n+1}$$ which is in the form $$\displaystyle \sum_{n=0}^\infty c_n x^{n+1}$$ Do you see the difference? The power of $x$ is off by one.
If you correct for that off-by-one error, you will get your new $c_0$ is $0$, your new $c_1$ is your old $c_0$, and so on. That is exactly the correct answer that you got the other way.
ADDED:
If you need a general formula for your new $c_n$ you could use
$$ c_n = \begin{cases} 0, & n=0 \\[2ex] \displaystyle \frac{(-1)^{n-1}}{5 \cdot 10^{n-1}}, & n>0 \end{cases} $$
Note that I changed the $n$'s to $n-1$'s in the formula for $c_n$ for this to work. It would be difficult to write that into an explicit series, as you did for $\sum_{n=0}^\infty \frac{(-1)^n}{5 \cdot 10^n} x^{n+1}$.
But you do not need such a formula, since you were only asked for $c_0$ through $c_4$.